Karl's Calculus Tutor - Solution to Exercise 7.2-3

Solution to Exercise 7.2-3

Taking this derivative can be done pretty much in one step. Let g(x) = x. Let h(x) = sin(x). Let u(x) = 1/x. Then

   g'(x)  =  1

   h'(x)  = cos(x)

            -1
   u'(x)  =   
            x²

If f(x) = g(x) h(u(x)), then combining the product rule and the chain rule will give you

   f'(x)  =  g'(x) h(u(x))  +  g(x) h'(u(x)) u'(x)

Now, substituting back the functions and derivatives for g(x), h(x), and u(x), we get

                                         -1
   f'(x)  =  1 × sin(1/x)  +  x cos(1/x)   
                                         x²

or simply

                          cos(1/x)
   f'(x)  =  sin(1/x)  -          
                             x

And what can you say about it at x = 0? Well, you can't take 1/x, and you can't divide cosine of it by zero. So you would have to say that this derivitative is undefined at x = 0. That is in contrast to the original function, f(x) = x sin(1/x). Although you still can't take 1/x at x = 0, you can find a limit of this function as x approaches zero. This is because, no matter how close to zero x is, -1 £ sin(1/x) £ 1. And so, no matter how close x is to zero, it must also be true that -|x| £ x sin(1/x) £ |x|. As x approaches zero, x sin(1/x) is squeezed between plus and minus x. And that squeezes the limit to exactly zero.

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